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\newcounter{ssection}
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\title{Basic Facts about Lie groups and its representation.}

\def\Ad{\mathrm{Ad}}
\def\agl{\mathrm{gl}}
\def\asl{\mathrm{sl}}
\def\aso{\mathrm{so}}
\def\asp{\mathrm{sp}}
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\def\SO{\mathrm{SO}}
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\begin{document}
\maketitle

\section{notation}
\begin{description}
\item[$B_K$] Killing form, i.e. 
  \[
  B_K(X,Y) = \tr(\ad X \ad Y)
  \]
\item[$B_T$] Trace form 
  \[
  B_T(X,Y) = \Re\tr(XY^*)
  \]
\item[$\Mat_n(F)$] set of matrix over field $F$;   
\item[$I_n$] identity matrix in $\Mat_n(\bC)$;
\item[$E_{k,l}$] matrix with only $(k,j)$-th entry $1$ and others are
  zero. 
\item[$e_l$] can be consider as an element in $\ft^*$ the dual space
  of cartan algebra.
\item[$H_\alpha$] coroot corresponding to $\alpha$, i.e. 
  \[
  B_K(H,H_\alpha) = \alpha(H).
  \]
\item[$\inn{\alpha}{\beta}$] 
  $\inn{\alpha}{\beta} = B_K(H_\alpha,H_\beta) = \alpha(H_\beta) = \beta(H_\alpha)$
\item[$\theta$] cartan involution of lie algebras
\item[$\Theta$] cartan involution of lie groups.
\end{description}

\section{Structures}
\subsection{basic facts}
Note that 
\[
E_{k,l}E_{s,t} = \delta_{l,s}E_{k,t}
\]
Then 
\[
[E_{s,s},E_{k,l}] = E_{s,s}E_{k,l} - E_{k,l}E_{s,s} 
= \delta_{s,k}E_{k,l} - \delta_{l,s}E_{k,l}
\]

\subsection{simple, semisimple, reductive groups}

\subsubsection{simple lie groups}

\subsubsection{semisimple lie groups}

\subsubsection{reductive group}
$\GL(n,\bC)$

\section{Complex lie group and its lie algebra}

\subsection{$\GL(n,\bC)$}
\[
\GL(n,\bC) = \Set{g\in \Mat_{n\times n}(\bC)|\det g \neq 0}
\]

Hence the Lie algebra is 
\[
\agl(n,\bC) = \Mat_{n\times n}(\bC)
\]

Note that 
\[
\agl(n,\bC) = \asl(n,\bC) \oplus Z_{\agl(n,\bC)}
\]
where 
\[
Z_{\agl(n,\bC)} = \bC I_n
\]

$\GL(n,\bC)$ is a complex lie group, with complex dimension $n^2$ and real
dimension $2n^2$.

\subsection{$\SL(n,\bC)$}
\[
\SL(n,\bC) = \Set{g\in \Mat_{n\times n}(\bC)|\deg g = 1}
\]

By $\det(\exp X ) = \exp(\tr X)$ 
\[
\asl(n,\bC) = \Set{g\in \Mat_{n\times n}(\bC)|\tr X = 0}
\]
it is a simple complex lie algebra of type $A_{n-1}$

$\SL(n,\bC)$ is a complex lie group, with complex dimension $n^2-1$,
real dimension $2(n^2-1)$.

The catan algebra is 
\[
\ft = \Set{X \in \asl(n,\bC) | X = \diag(a_1, \cdots, a_n)}
\]
correspond $T$ is all diagonal matrix wih determinat $1$. 

Now, choose basis of $\ft$ be $H_l = iE_{l,l}$, $l = 1,\cdots, n$.
\[
[H_l, E_{l,s}] = i E_{l,s}, \quad l\neq s
\]
\[
[H_s, E_{l,s}] = -iE_{l,s}, \quad l\neq s
\]
Hence 
The nontrival roots are $e_l - e_s$ with root sapces
$E_{l,s}$ ($l\neq s$).

Choose a \index{positive root system} be $e_l-e_s$ positive iff $l<s$.
Corresponding standard\index{simple roots} are $e_l-e_{l+1}$ for $l=1, \cdots, n-1$. Call it $A_{n-1}$.

Since $\asl(n,\bC)$ is a simple complex lie algebra. 
The invariant bilinear form is 
%%need an reference to sec:bilinear_form_simple
unique up to scale.

The relation between killing form over $\asl(n,\bC)$ and trace form is 
\[
2n\,B_T(X,Y) = B_K(X,Y).
\]
This can be calculated by let $X=Y=E_{1,1}-E_{2,2}$.

Let $H_{st} = E_{s,s} - E_{t,t}$.

Then 
\[
H_{e_l-e_k} = \frac{1}{2n} (E_{l,l}-E_{k,k}).
\]

In fact,
\[
(e_l-e_k)(H_{st}) =\delta_{ls} - \delta_{lt} - \delta_{ks} +\delta_{kt}=
B_T(H_{lk}H_{st}).
\]

Then 
\[
\inn{e_l-e_k}{e_s-e_t} 
= \frac{1}{4n^2}(\delta_{ls} - \delta_{lt} - \delta_{ks} + \delta_{kt}).
\]

\[
\frac{2\inn{e_l-e_{l+1}}{e_k-e_{k+1}}}{\inn{e_k-e_{k+1}}{e_k-e_{k+1}}} =
\begin{cases}
  2 & l=k \\
  -1 & l = k\pm 1
\end{cases} 
\]
 
Cartan matrix respect to the standard simple root is 
\[
\begin{pmatrix}
  2  & -1 &  0 & 0 & \cdots & 0 \\
  -1 &  2 & -1 & 0 & \cdots & 0 \\
  0  & -1 &  2 & -1& \cdots & 0 \\
  \cdots & \cdots & \cdots & \cdots & \cdots &\cdots \\
  0 & 0 & 0 & 0 & \cdots & 2 
\end{pmatrix}
\] 

The borel subalgebra is upper triangle matrices in $\asl(n,\bC)$. 

\subsection{$\SO(2n,\bC)$ and $\SO(2n+1,\bC)$}
\[
\SO(n,\bC) = \Set{g\in \GL(n,\bC)| g^Tg = 1}
\]
Correspond lie algebra
\[
\aso(n,\bC) = \Set{X\in \agl(n,\bC)| X^T + X = 0}
\]

The diagnal elements are all zero for any $X\in \aso(n,\bC)$.
Hence the complex dimension of the lie algebra is 
$n(n-1)/2$.

Choose the cartan subalgebra be 
\[
\ft = \begin{pmatrix}
   0 & h_1 &   0 & 0 & \cdots & 0 & 0 & 0 \\
-h_1 & 0   &   0 & 0 & \cdots & 0 & 0 & 0 \\
   0 & 0   &   0 & h_2 & \cdots & 0 & 0 & 0 \\
   0 & 0   & -h_2&  0 & \cdots & 0 & 0 & 0 \\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
   0 & 0   & 0   &  0 & \cdots & 0 & h_n & 0 \\
   0 & 0   & 0   &  0 & \cdots & -h_n& 0 & 0 \\
   0 & 0   & 0   &  0 & \cdots & 0 & 0 & 0 
\end{pmatrix}
\]
for $\aso(2n+1,\bC)$ and delete the last row and column for $\aso(2n,\bC)$.

\subsubsection{$\aso(2n,\bC)$}
Consider the block matrix, with $k,l$-the block:
\[
Y = \begin{pmatrix}
  0  & X \\
  -X^T & 0  
\end{pmatrix}
\]
where $X\in \Mat_{2\times 2}(\bC)$.

\[
S = \begin{pmatrix}
   0 & h_k &   0 & 0 \\
-h_k & 0   &   0 & 0 \\
   0 & 0   &   0 & h_l \\
   0 & 0   & -h_l&  0 
\end{pmatrix}
\]
By easy caculation we see that:
\[
\begin{cases}
[S,Y] = (ih_k+ih_l) Y
& X = \begin{pmatrix}
  1 & i \\
  i & -1
\end{pmatrix} \\
[S,Y] = (ih_k-ih_l) Y
& X = \begin{pmatrix}
  1 & -i \\
  i & 1
\end{pmatrix} \\
[S,Y] = (-ih_k-ih_l) Y
& X = \begin{pmatrix}
  -1 & i \\
  i & 1
\end{pmatrix} \\
[S,Y] = (-ih_k+ih_l) Y
& X = \begin{pmatrix}
   1 & i \\
  -i & 1
\end{pmatrix} 
\end{cases}
\]

Let $e_k(H) = -ih_k$, then all the nontrivial roots are $\pm e_k \pm e_l$.

Still choose the positive system by \index{lexticographic order}
\subsubsection{$\aso(2n+1, \bC)$}

Consider the last column now. Let
\[
S = \begin{pmatrix}
  0 & h_l & 0 \\
  -h_l & 0 & 0 \\
  0 & 0 &0 
\end{pmatrix}
\]

\[
Y = \begin{pmatrix}
  0 & X\\
  -X^T & 0
\end{pmatrix}
\]
where $X\in \Mat_{2\times 1}(\bC)$.
Then 
\[
\begin{cases}
[S,Y] = -ih_k Y & 
X = \begin{pmatrix}
  1 \\
  -i
\end{pmatrix} \\
[S,Y] = ih_k Y & 
X = \begin{pmatrix}
  1 \\
  i
\end{pmatrix} 
\end{cases}
\]
Hence $\pm e_k$ are extra roots for $\aso(2n+1,\bC)$ than $\aso(2n,\bC)$.



\subsection{$\Sp(n,\bC)$}
Let 
\[
J = \begin{pmatrix}
0 & I_n \\
-I_n & 0
\end{pmatrix}
\]
Define
\[
\Sp(n,\bC) = \Set{g\in \GL(2n,\bC)| g^TJg = J}.
\]

The lie algebra is 
\[
\asp(n,\bC) = \Set{X\in \Mat_{2n\times 2n}(\bC)| X^TJ + JX = 0}.
\]

If we write 
\[
X = \begin{pmatrix}
  A & B \\
  C & D 
\end{pmatrix},
\]
by above equation, we have
\begin{align*}
X^TJ = & 
\begin{pmatrix}
  A^T & C^T \\
  B^T & D^T
\end{pmatrix}
\begin{pmatrix}
  0 & I_n \\
  -I_n & 0
\end{pmatrix}
= \begin{pmatrix}
-C^T & A^T \\
 -D^T & B^T
\end{pmatrix}\\
JX = & 
\begin{pmatrix}
  0 & I_n \\
  -I_n & 0
\end{pmatrix}
\begin{pmatrix}
  A & B \\
  C & D
\end{pmatrix}
= \begin{pmatrix}
 C & D \\
 -A & -B
\end{pmatrix}
\end{align*}

Hence 
\begin{align*}
D =& -A^T\\
B =& B^T \\
C =&  C^T  
\end{align*}

Hence the complex dimension of the lie algebra and lie group is 
$n^2+n(n+1) = n(2n+1)$.

The typical Cartan subalgebra is 
\[
\begin{split}
\ft &=\sspan\Set{E_{l,l}-E_{n+l,n+l}|l=1,\cdots, n} \\
&= 
\Set{\begin{pmatrix}
    h_1   &   0   & \cdots&  0   & 0    &   0  &\cdots& 0 \\
    0     &   h_2 & \cdots&  0   & 0    &   0  &\cdots& 0  \\
    \cdots& \cdots& \cdots&\cdots&\cdots&\cdots&\cdots& \cdots \\
    0     &   0   & \cdots&  h_n & 0    &   0  &\cdots& 0 \\
    0     &   0   & \cdots&  0   & -h_1 &   0  &\cdots& 0 \\
    0     &   0   & \cdots&  0   &   0  &-h_2  &\cdots& 0 \\
    \cdots& \cdots& \cdots&\cdots&\cdots&\cdots&\cdots& \cdots \\
    0     &   0   &\cdots &  0   &    0 &   0  &\cdots& -h_n     
  \end{pmatrix}}.
\end{split}
\]

Let 
\[
e_l(X) = h_l,
\]
where $X$ is in above form.

Now consider 
\[ S = 
 \begin{pmatrix}
  H & 0 \\
  0 & -H
\end{pmatrix}
\]
in $\ft$, where $H$ is a diagnal matrix.
Let 
\[
X = \begin{pmatrix}
  A & B \\
  C & -A^T
\end{pmatrix}
\] be an element in $\asp(2n,\bC)$, where $B, C$ are symetric matrix. 

By elementary cumputation, we can see
\[
[S, X] = 
\begin{pmatrix}
  HA-AH & HB + BH \\
  -(HC+CH) & -(HA-AH)^T 
\end{pmatrix}
\]

Then we can see the roots and corresponding root vectors are
\begin{align*}
e_l-e_k : & E_{l,k} -E_{k+n,l+n} \\
e_l+e_k : & E_{l, k+n} + E_{k,l+n} \\
-(e_l+e_k) :& E_{l+n,k} + E_{k+n,l} ,
\end{align*}
where $l,k = 1, \cdots, n$.

Choose positive system $\Delta=\Set{e_l-e_k|l<k} \cup \Set{e_l+e_k}$.
Then the set of simple roots is $\Set{e_l-e_{l+1}|l=1,\cdots, n} \cup \Set{2e_{n}}$.

Let $S = E_{1,1} - E_{1+n,1+n}$.
Then 
\[
B_K(S,S) = \sum_{\alpha\in \Delta} \alpha(S)^2 = 2*(n-1) + 4*n = (3n-1) B_T(S,S)
\]

we can choose (under $B = \frac{1}{2}B_T$)
\[
H_{\pm e_l \pm e_k} = \pm(E_{l,l}-E_{l+n,l+n}) \pm (E_{k,k}-E_{k+n,k+n})
\]

Then 
\[
\frac{2\inn{e_l-e_k}{e_s-e_t}}{\inn{e_l-e_k}{e_l-e_k}}
=\delta_{l,s} + \delta_{k,t}-\delta_{k,s} -\delta_{l,t}
\]
\[
\frac{2\inn{e_l-e_k}{e_s+e_t}}{\inn{e_l-e_k}{e_s+e_k}}
=\delta_{l,s} + \delta_{l,t}-\delta_{k,s} -\delta_{k,t}
\]

The cartan matrix is 
\[
\begin{pmatrix}
 2 & -1 &  0 & \cdots & 0 & 0 \\
-1 &  2 & -1 & \cdots & 0 & 0 \\
0  & -1 &  2 & \cdots & 0 & 0 \\
\cdots & \cdots &\cdots & \cdots &\cdots & \cdots \\
0  & 0  &  0 & \cdots & 2 & -2 \\
0  & 0  &  0 & \cdots & -1 & 2 
\end{pmatrix}
\]


\section{Real Lie groups and its algebra}

\subsection{$\GL(n,\bR)$}
\[
\GL(n,\bR) = \GL(n,\bC) \cap \Mat_{n\times n}(\bR)
\]
Consider the complex conjugate $\sigma$ of $\agl(n,\bC)$, 
i.e $\sigma(X) = \overline{X}$. It is a involution on $\agl(n,\bC)$.
$\agl(n,\bR)$ is the set of fix points of $\sigma$.

Then one cartan involution of $\agl(n,\bR)$ is $\theta(X) = -X^*$,
which commutes with $\sigma$.

Now 
\[ 
\begin{split}
B_\theta(X,Y) &= -B_K(X, \theta(Y)) 
\end{split}
\]

\subsection{$\SL(n,\bR)$}
\[
\SL(n,\bR) = \SL(n,\bC) \cap \GL(n,\bR)
\]



\subsection{$U(n)$}
Consider the involution $\theta$ of $\agl(n,\bR)$:
\[
\theta(X) = -X^*
\]
$\au(n)$ is the set of fixed point of $\theta$.

\[
\U(n) = \Set{g \in \GL(n,\bC)|g^*g = I}
\]

Lie algebra: 

\[
u(n,\bC) = \Set{X\in \GL(n,\bC)|X^* + X = 0}
\]

Then the real dimension of $U(n)$ is $2\frac{(n-1)n}{2}+ n = n^2+1$. 
It is simpliy connected
when $n>1$. When $n=1$ it is torus, hence connected with fundermantal
group $\bZ$.

The typical cartan subalgebra is 
\[
\ft = \Set{X\in \Mat_{n\times n}(\bC)| X = \diag{a_1, \cdots, a_n},
  a_l\in i\bR}
\] 
corresponding maximal torus $T$ of $U(n,\bC)$ is the diagnal matrics with
entries on the torus $S^1$. Clearly $T\cong S^n$.



The roots correspond to $T$ spaned by  
\[
\Set{e_l}
\]
Where 

Then the nontrivial roots are $e_l-e_s$ and $-(e_l-e_s)$for any $l\neq s$, 
with root vectors $ d$ and $(1-i)E_{ls}$


\section{Symplectic groups}
\subsection{polarizations}

Let $W$ be a symplectic space, $W = U \oplus U'$ be a complete polization.
Let $P(U) = \Stab_{\Sp(W)} U$ be the stablizer of space $U$. 
Then there is a natural map 
\[
r\colon P(U) \to \GL(U)
\]

Let $N(U,W) = N(U)=N$ be the kernel of map $r$. 


\subsection{Wallach}
assume $W_\bC=W\otimes \bC$, where $W_0$ is a symplectic space over $\bR$. 
$W$ have a symplectic structrue over $\bC$ extend from $W_0$, 
define $\sigma$ be the conjugation on $\bC$ extend to $W$. 
Then we can define a real form $\Sp(W)$ 
of $\Sp(W)$ by the commutator of $\sigma$.
Moreover, it is clear this act on the eigen space of $\sigma$ which are $W$
and $W\otimes i$.

suppose that $W= U\oplus U'$ be a compelet polization. 
define $\theta$ be multiply $i$ on $U\otimes \bC$ and $-i$ on $U'\otimes \bC$. 
(or multiply $1$, $-1$ on $U$ and $U'$.)
this gives a cartan involution on $\gg=\sp(W)$ or on $\gg_\bC$.
$\kk= \Set{X\in \gg|X\circ \theta = \theta X}$
is the commutator of $\theta$ and
$\pp=\Set{X\in gg|X\circ \theta = - \theta X}$ 
is the anti-commutator of $\theta$.
moreover $X$ should act on the eigen space of $\theta$, which are
$U$ and $U'$ hence in fact $K = P(U)\cap P(U') = U(n)$ 

Let $W=W_1\oplus W_2$, where $\dim W_1 = 2r$, $\dim W_2 = 2s$. 


\section{bilinear form}
\subsection{invariant form over lie algebra}
Invariant bilinear form $B$ over simple lie algebra $\fg$ is a bilinear 
form satifies 
\[
B(\ad(Z)X,Y) =  -B(X,\ad(Z)Y), \quad X,Y\in \fg
\]

If $\fg$ is a lie algebra of some lie group $G$, then 
\[
B(\Ad(g)X, \Ad(g)Y) = B(X,Y), \quad g\in G, X,Y\in \fg
\]

\subsection{killing form}
Define \index{killing form} by 
\[
B_K(X,Y) = \tr(\ad X \ad Y), \quad X,Y\in \fg
\]

\subsection{trace form}
For real linear lie algebra,
define \index{trace form}
\[
B_T(X,Y) = \tr(XY).
\]
clearly, it is a symetric bilnear form (by compare the explicit expressions of
both sides), i.e. 
\[
B_T(X,Y) = B_T(Y,X)
\]
moreover 
\[
\begin{split}
B_T(\ad(Z)X,Y) &= \tr([Z,X]Y) = \tr(ZXY-XZY) \\
&=  \tr(XYZ - XZY) =
\tr(X[Y,Z]) = - B_T(X, \ad(Z)Y)
\end{split}
\]

\subsection{bilinear form over simple lie algebra} 
\label{sec:bilinear_form_simple}
The radical $\rad(B) = \Set{Y\in \fg|B(X,Y) = 0\, \forall X\in \fg}$ 
of the invariant bilinear form $B$ is a ideal of lie algebra
$\fg$. 
In fact, if $Y\in \rad(B)$, 
\[
B(X,\ad(Z) Y) = -B(\ad(Z)X,Y) = 0.
\]
Hence $\ad(Z) Y \in \rad(B)$. 

So, if the lie algebra is simple, nontrival invariant bilinear form
is nondegenerate. 

In fact, the bilinear form give a hermitian structure on the adjoint
representation of $\fg$. $\fg$ is simple lie algebra means this
representation is irreducible. Thus the bilinear form is unique up to scaler.

\section{Index}

\end{document}

